Question

Let \(X, Y, Z\) be Banach spaces, let \(T\) be a bounded linear operator from \(X\) into \(Y\) such that \(T(X)\) is closed in \(Y\), and let \(S\) be a finite-rank operator from \(X\) into \(Z\) (that is, \(\operatorname{dim}(S(X))\) is finite). Define \(U: X \rightarrow Y \oplus Z\) by \(U(x)=(T(x), S(x))\). Show that \(U(X)\) is closed in \(Y \oplus Z\)

Short answer

Since \(T(X)\) is closed in \(Y\) and \(S(X)\) is finite-dimensional, \(U(X)\) is closed in \(Y \oplus Z\) by properties of closed and finite-dimensional sets.

Step-by-step solution

- Define the Direct Sum

Recall that the direct sum of two Banach spaces, denoted by \(Y \bigoplus Z\), is the space of pairs of elements from each space, i.e., \((y, z)\) for \( y \in Y \) and \( z \in Z \).

- Construct Operator U

Given \(T\) and \(S\), define the operator \(U: X \rightarrow Y \bigoplus Z\) by \(U(x) = (T(x), S(x))\) for all \(x \in X\).

- Characterize the Image of U

Observe that \( U(X) = \{(T(x), S(x)) \in Y \bigoplus Z : x \in X\}\). Note that \(T(X)\) is closed in \(Y\) and \(S(X)\) is finite-dimensional.

- Use Properties of Closed Sets and Finite Dimensions

Since \(T(X)\) is closed and \(S(X)\) is finite-dimensional, \(S(X)\) is also closed in \(Z\) by the property that finite-dimensional subspaces of Banach spaces are closed.

- Prove U(X) is Closed

By the definition of the direct sum, the image \(U(X)\) is closed in \(Y \bigoplus Z\) since it is the product of closed sets \(T(X)\) and \(S(X)\). Therefore, \(U(X)\) is closed in \(Y \bigoplus Z\).

Key concepts

bounded linear operator

In Banach spaces, a bounded linear operator is a type of linear map between two Banach spaces where the size of the output is controlled by a fixed multiple of the size of the input. Formally, let \(X\) and \(Y\) be Banach spaces, and let \(T\) be a linear operator from \(X\) to \(Y\). Then, \(T\) is said to be bounded if there exists a constant \(C\) such that for all \(x \in X\), \|T(x)\| \leq C \|x\|. This ensures that \(T\) is continuous.

Bounded linear operators play a crucial role in functional analysis due to their predictable behavior. For example, in our exercise, \(T\) is a bounded linear operator which helps us to define another operator \(U\) composed of \(T\) and another operator \(S\). The boundedness of \(T\) ensures that \(U\) will behave well when mapping from one Banach space to another.

finite-rank operator

A finite-rank operator is a special type of bounded linear operator where the image (or range) has finite dimensions. Let \(X\) and \(Z\) be Banach spaces, and \(S\) be a linear operator from \(X\) to \(Z\). \(S\) is a finite-rank operator if the dimension of \(S(X)\), the image of \(S\), is finite. More formally, this means that there exists a finite-dimensional subspace \(F \,\subseteq\, Z\) such that \(S(X) \subseteq F\).

These operators are significant because their properties are easier to handle compared to operators with infinite-dimensional ranges. For instance, in the exercise, the finite-rank nature of \(S\) guarantees that \(S(X)\) is closed in \(Z\). This property is used to show that the combined operator \(U\) has a closed image in the direct sum space, \(Y \bigoplus Z\).

direct sum

The direct sum of two Banach spaces \(Y\) and \(Z\), denoted by \(Y \oplus Z\), is a new Banach space that pairs each element from \(Y\) with each element from \(Z\). Formally, \(Y \oplus Z\) consists of all ordered pairs \((y, z)\) where \(y \in Y\) and \(z \in Z\). This new space inherits properties from both \(Y\) and \(Z\) and is typically equipped with a norm defined by \(\|(y, z)\| = \|y\| + \|z\|\).

The direct sum is a powerful concept as it allows combining spaces and operators to form new ones. In our exercise, we define the operator \(U(x) = (T(x), S(x))\) which maps to \(Y \oplus Z\). Understanding the direct sum helps us comprehend how \(U\) functions and why \(U(X)\) being closed in \(Y \oplus Z\) follows from the properties of \(T\) and \(S\).

closed subspace

A subspace is a subset of a Banach space that is itself a Banach space with the inherited operations and norms. A closed subspace is one where if a sequence of points within the subspace converges to a limit point, that limit point also lies within the subspace. This is a critical concept because many results in functional analysis rely on the completeness of these subspaces.

For example, in our exercise, both \(T(X)\) and \(S(X)\) are closed subspaces of \(Y\) and \(Z\), respectively. This property ensures that their direct sum, \(U(X)\), is closed in \(Y \oplus Z\). Closed subspaces are vital in proving that operators like \(U\) have well-behaved images, allowing us to apply various theorems from topology and functional analysis to infer further properties about \(U(X)\).